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(D) For the block to be in equilibrium,the upward frictional force $f$ must balance the downward gravitational force $mg$. Thus,$f = mg$.The maximum s

Vedclass · Accepted Answer

The minimum spring constant $k_{\min}$ to keep the block of mass $m$ in equilibrium is $\frac{mg}{\mu d}$.

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(D) For the block to be in equilibrium,the upward frictional force $f$ must balance the downward gravitational force $mg$. Thus,$f = mg$.The maximum static frictional force is given by $f_{\max} = \mu N$,where $N$ is the normal reaction force exerted by the wall on the block.The spring is compressed by a distance $d$,so the normal force exerted by the spring on the block is $N = kd$.For equilibrium,the frictional force must satisfy $f \leq f_{\max}$,which implies $mg \leq \mu N$.Substituting $N = kd$,we get $mg \leq \mu kd$.Rearranging for $k$,we find $k \geq \frac{mg}{\mu d}$.Therefore,the minimum spring constant required to keep the block in equilibrium is $k_{\min} = \frac{mg}{\mu d}$.

$A$ block of mass $m$ is pressed against a vertical surface by a spring of unstretched length $l$. The spring is compressed by a distance $d$ as shown in the figure. If the coefficient of friction between the block and the surface is $\mu$,choose the correct statement.

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